The acceleration-time graph for a particle mo | vedclass

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Question

(a)

The area under $a-t$ graph gives change in velocity.

Given, $u=-5 \,m / s$

$\Rightarrow$ Area on positive side $=\frac{1}{2} 	imes 6 	i

Vedclass · Accepted Answer

$+15$

Vedclass · Answer

(a)

The area under $a-t$ graph gives change in velocity.

Given, $u=-5 \,m / s$

$\Rightarrow$ Area on positive side $=\frac{1}{2} 	imes 6 	imes 10=30 \,ms ^{-1}$

$\Rightarrow$ Area on negative side $=\frac{1}{2} 	imes 2 	imes 10=10 \,ms ^{-1}$

Net area $=30-10=20 \,ms ^{-1}$

$\Delta v=$ Area

$v-(-5)=20$

$\Rightarrow v=15 \,ms ^{-1}$

Graph	Characteristics
$(A)$	$(i)$ has $v > 0$ and $a < 0$ throughout
$(B)$	$(ii)$ has $x > 0,$ throughout and has a point with $v = 0$ and a point with $a = 0$
$(C)$	$(iii)$ has a point with zero displacement for $t > 0$
$(D)$	$(iv)$ has $v < 0$ and $a > 0$

The acceleration-time graph for a particle moving along $x$-axis is shown in figure. If the initial velocity of particle is $-5 \,m / s$, the velocity at $t=8 \,s$ is ....... $m / s$

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